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We have fU(X)f(_) = (3) fi(X)2 Proof: We first multiply Equation (1) by the denominator of its right hand side and then differentiating both side with respect to x. We get 2f(x)f'(r)(Rd + fr 1dt) T y = eCIC2 = ae where a > 0 and b > 0. (Note that b > 0 follows from the fact that f is decreasing. ) In order to determine a and b, we substitute f (r) = ae- bx into Equation (4) and check the two boundary points x - 0 and x = L. We obtain the following two equations: -2rof(x). = fJ0 f (t) Iny, + C2 = cobRda + roco(e cobRda + roco(e Since f(x) 0 0, we can divide both side by f(x) and get )a - rebCL - 1)a - - robCLe = 0, = 0.

Hence the total signal delay D = ,3=, Di. We represent the wire resistance (capacitance) of Q, by Ri (Ci). We have R2 = r and C2 = cody. The signal delay through the wire can be calculated as follows: roAx -T 3 f)f(t) f rx, lx X R d f( { Thus dD = coJ(Rd + ro f(t)dt) y2 By setting dD = 0, we get 2 We now consider unconstrained wire-sizing. We show that the optimal wire-sizing function satisfies a second order ordinary differential equation which can be analytically solved. - dt)- rO 5(CL + Co :+ Unconstrained Wire-Sizing ire J-21 10 ro(CL + co fag f(t)dt) co(Rd + rO jo+ 7 dt) ym-n gives minimum delay.

Section III will show an example for which no track permutations are possible, while track segment permutations lead to optimal results. In [10], track segment permutations in a switchbox are considered. In Section III, we show that even this method is sub-optimal because the initial route is not driven by coupled noise considerations. Besides, drive strengths are not considered and Section II will show that the coupled noise model is simplistic. The channel routing heuristic presented in [11] modifies a well-known channel router [14], [15] to reflect coupled noise issues.

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